# B My Second Post: Change in Magnitude of Normal Force?

#### Weam Abou Hamdan

Hello,

Let us imagine a solid immersed in a liquid in a container such that the density of the liquid is less than the density of the solid. This means that the solid must sink. Let us study the solid when it reaches the bottom surface and is now at rest. The forces acting on the solid when it is at the surface are its Weight (W), the Normal Force (N) exerted by the bottom surface, and the Buoyant Force (FB) exerted by the liquid. According to Newton's First Law of Motion, since the solid is at equilibrium, the sum of these forces must be zero.
W = FB + N
Let us now study the same solid resting on the same bottom surface of the container such that there is no liquid. The forces acting on the solid in this case are its Weight and the Normal Force (N2) exerted by the same surface. According to Newton's First Law of Motion, since the solid is at equilibrium, the sum of the forces must be zero.
W=N2
Apparently, the magnitude of the Normal Force increased from the first case to the second case. What is the physical origin of such an increase?

Weam Abou Hamdan
Thursday, July 19, 2018

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#### jbriggs444

Homework Helper
The forces acting on the solid when it is at the surface are its Weight (W), the Normal Force (N) exerted by the bottom surface, and the Buoyant Force (FB)
Buoyant force is nothing more or less than the difference in vertical force resulting from the pressure difference between an object's upper and lower surfaces.

If you are counting both the force from the bottom surface and the net force from buoyancy, you are likely doing an improper accounting for forces.

#### Weam Abou Hamdan

If you are counting both the force from the bottom surface and the net force from buoyancy, you are likely doing an improper accounting for forces.
Are you saying that, since the solid is in contact with the bottom surface, then pressure from the liquid only exists on the upper surface?

#### Doc Al

Mentor
What is the physical origin of such an increase?
In the first case one assumes at least a small layer of fluid beneath the object, otherwise there will be no buoyant force.

#### Weam Abou Hamdan

In the first case one assumes at least a small layer of fluid beneath the object, otherwise there will be no buoyant force.
How is the bottom surface able to exert a Normal Force on the solid, then?

#### Doc Al

Mentor
How is the bottom surface able to exert a Normal Force on the solid, then?
The surface isn't perfectly flat. If it is, and there is a tight seal, then you'll have no upward force from the fluid.

#### Weam Abou Hamdan

The surface isn't perfectly flat. If it is, and there is a tight seal, then you'll have no upward force from the fluid.
What I mean is how can the solid experience both a Normal Force and the Buoyancy Force if there can only be one material below the solid (either the bottom surface or the liquid)?

#### Doc Al

Mentor
What I mean is how can the solid experience both a Normal Force and the Buoyancy Force if there can only be one material below the solid (either the bottom surface or the liquid)?
I see what you're saying, but imagine the surfaces at the micro-level being quite jagged. So both fluid and floor can exert forces.

But if the seal with the floor is tight, that's a different story.

#### Weam Abou Hamdan

In the first case one assumes at least a small layer of fluid beneath the object, otherwise there will be no buoyant force.
In that case, that should mean there is no Normal Force because there is no contact. However, there is a Normal Force.

#### russ_watters

Mentor
What I mean is how can the solid experience both a Normal Force and the Buoyancy Force if there can only be one material below the solid (either the bottom surface or the liquid)?
In real life there can be two.

#### Weam Abou Hamdan

In real life there can be two.
It doesn't really make sense though. There's the solid particle. Below it is the liquid particle. Below that is the container particle.

#### russ_watters

Mentor
It doesn't really make sense though. There's the solid particle. Below it is the liquid particle. Below that is the container particle.
What "particle"? You said "solid". Are you looking for a zero size point particle now?

You do realize no real object is perfectly flat and smooth, right?

#### Weam Abou Hamdan

What "particle"? You said "solid". Are you looking for a zero size point particle now?

You do realize no real object is perfectly flat and smooth, right?
I did say "solid". What i meant is one of the particles on the bottom part of the solid.
Replacing the solid with a zero size point particle would be effective as well.

#### russ_watters

Mentor
I did say "solid". What i meant is one of the particles on the bottom part of the solid.
Replacing the solid with a zero size point particle would be effective as well.
If you have a zero size point particle it has no mass or surface area, so you can't do the problem with it. Again:

You do realize no real object is perfectly flat and smooth, right?

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There are at least a couple of possible scenarios in this problem, and the result is different for both. With a water tight seal at the bottom, the buoyant force from the water may be completely absent. Alternatively, if enough water can get in the space between the bottom and the solid, the full buoyant force as given by Archimedes principle will occur.

#### Weam Abou Hamdan

There are at least a couple of possible scenarios in this problem, and the result is different for both. With a water tight seal at the bottom, the buoyant force from the water may be completely absent. Alternatively, if enough water can get in the space between the bottom and the solid, the full buoyant force as given by Archimedes principle will occur.
In the second case where the water can get in the space, is there a Normal Force exerted by the bottom surface? If so, how?

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In the second case where the water can get in the space, is there a Normal Force exerted by the bottom surface? If so, how?
Suggestion would be to place the object on 4 thin legs, and measure the force that the 4 thin legs supply. You need to have a well-defined scenario. A bottom consisting of a bunch of pebbles or sand is rather poorly defined, and difficult to determine how much of the upward force would be supplied by the buoyant force from the water.

#### russ_watters

Mentor
A bottom consisting of a bunch of pebbles or sand is rather poorly defined, and difficult to determine how much of the upward force would be supplied by the buoyant force from the water.
This can be tested though and I'd bet that there is zero or near zero measurable deviation for a real object. This is because a hard object has a tiny contact area and a soft object traps water and maintains the pressure.

Also I feel like there should be a conservation law reason why the forces can't change on their own. E.G., to eliminate buoyancy you would need a pump, not just the weight of the object.

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This can be tested though and I'd bet that there is zero or near zero measurable deviation for a real object. This is because a hard object has a tiny contact area and a soft object traps water and maintains the pressure.

Also I feel like there should be a conservation law reason why the forces can't change on their own. E.G., to eliminate buoyancy you would need a pump, not just the weight of the object.
Post 54 of this thread https://www.physicsforums.com/threads/ball-floating-on-the-surface-of-water.944587/page-3#post-5977539 is when I first took note of what can be an exception to Archimedes principle. The precise conditions for which this exception would apply are open to discussion. In the case of a flat solid that rests on a couple of bricks with flat surfaces, I would expect the area of the bricks would reduce the product of $F_{buoyant}= PA$ by reducing the area $A$. $\\$ How much buoyant force is supplied by the water is most readily determined by a cable that is attached above the object that attempts to lift the object. If the water fails to supply the complete buoyant force as specified by Archimedes, this would be determined as the cable first starts to pull on the object, before it starts to lift it.

Last edited:

#### nasu

In the second case where the water can get in the space, is there a Normal Force exerted by the bottom surface? If so, how?
There is a normal force exerted by the bottom surface in any case. Even before you deep the solid, there is the normal force exerted by the bottom surface on the liquid. When you put the solid in, the normal force will be exerted on the solid in the points where you have contact solid-bottom and on the liquid in the areas where there is a layer of liquid in between the two solids.
So the normal force is always exerted by the bottom surface. What may be different is on what that force is exerted.

#### jbriggs444

Homework Helper
In the second case where the water can get in the space, is there a Normal Force exerted by the bottom surface? If so, how?
It is indeterminate.

We do not know how much of the space in between has water and how much has direct contact. But as long as the block is at rest on the bottom, it does not matter. The sum of the normal force and the force from water pressure on the bottom must sum with all of the other forces acting to just enough to keep the block at rest. Otherwise it would not be at rest.

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A suction cup may be one of the better examples of how the full buoyant force may not apply if the pressure from the air or water can not reach the bottom surface. The suction cup may be an extreme case, but it illustrates the concept.

#### OldYat47

The issue you are overlooking is the difficulty squeezing out a very thin liquid layer. As the liquid layer thins out there is still hydraulic pressure in the gap but the liquid resists moving, transferring some of the vertical force directly to the bottom of the container. The liquid layer resists squeezing out because of the affinity of the individual molecules.

"My Second Post: Change in Magnitude of Normal Force?"

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